Match the items of column $-I$ with the items of column $-II$.
Column $-I$ Column $-II$ $(A)$ Methanol $(1)$ Conversion of phenol to $o-$hydroxybenzoic acid $(B)$ Kolbe's reaction $(2)$ Ethyl alcohol $(C)$ Williamson's synthesis $(3)$ Conversion of phenol to salicylaldehyde $(D)$ Conversion of $2^o-$ alcohol to ketone $(4)$ Wood spirit $(E)$ Reimer-Tiemann reaction $(5)$ Heated copper at $573 \ K$ $(F)$ Fermentation $(6)$ Reaction of alkyl halide with sodium alkoxide
| Column $-I$ | Column $-II$ |
|---|---|
| $(A)$ Methanol | $(1)$ Conversion of phenol to $o-$hydroxybenzoic acid |
| $(B)$ Kolbe's reaction | $(2)$ Ethyl alcohol |
| $(C)$ Williamson's synthesis | $(3)$ Conversion of phenol to salicylaldehyde |
| $(D)$ Conversion of $2^o-$ alcohol to ketone | $(4)$ Wood spirit |
| $(E)$ Reimer-Tiemann reaction | $(5)$ Heated copper at $573 \ K$ |
| $(F)$ Fermentation | $(6)$ Reaction of alkyl halide with sodium alkoxide |
(A-4, B-1, C-6, D-5, E-3, F-2) The correct matches are:
$(A)$ Methanol is known as Wood spirit $(4)$.
$(B)$ Kolbe's reaction is the conversion of phenol to $o-$hydroxybenzoic acid (salicylic acid) $(1)$.
$(C)$ Williamson's synthesis involves the reaction of alkyl halide with sodium alkoxide $(6)$.
$(D)$ Conversion of $2^o-$ alcohol to ketone is achieved by using heated copper at $573 \ K$ $(5)$.
$(E)$ Reimer-Tiemann reaction is the conversion of phenol to salicylaldehyde $(3)$.
$(F)$ Fermentation is the process used to produce Ethyl alcohol $(2)$.
Thus,the correct sequence is: $(A-4, B-1, C-6, D-5, E-3, F-2)$.
$(A)$ Methanol is known as Wood spirit $(4)$.
$(B)$ Kolbe's reaction is the conversion of phenol to $o-$hydroxybenzoic acid (salicylic acid) $(1)$.
$(C)$ Williamson's synthesis involves the reaction of alkyl halide with sodium alkoxide $(6)$.
$(D)$ Conversion of $2^o-$ alcohol to ketone is achieved by using heated copper at $573 \ K$ $(5)$.
$(E)$ Reimer-Tiemann reaction is the conversion of phenol to salicylaldehyde $(3)$.
$(F)$ Fermentation is the process used to produce Ethyl alcohol $(2)$.
Thus,the correct sequence is: $(A-4, B-1, C-6, D-5, E-3, F-2)$.
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$p-$Cresol reacts with chloroform in an alkaline medium to give compound $A$,which reacts with hydrogen cyanide to form compound $Y$. Hydrolysis of $Y$ gives a chiral carboxylic acid. Identify the structure of the carboxylic acid.
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DifficultTS EAMCET 2021
View SolutionMatch the compounds in Column-$I$ with their $C-O$ bond length in Column-$II$ and $C-O-H$ bond angle in Column-$III$.
Column-$I$ (Name)
Column-$II$ ($C-O$ bond length) and Column-$III$ ($C-O-H$ angle)
$A$. Methanol
$1$. $141 \ pm$,$i$. $109^o$
$B$. Phenol
$2$. $136 \ pm$,$ii$. $111.7^o$
$C$. Methoxymethane
$3$. $142 \ pm$,$iii$. $108.9^o$
| Column-$I$ (Name) | Column-$II$ ($C-O$ bond length) and Column-$III$ ($C-O-H$ angle) |
|---|---|
| $A$. Methanol | $1$. $141 \ pm$,$i$. $109^o$ |
| $B$. Phenol | $2$. $136 \ pm$,$ii$. $111.7^o$ |
| $C$. Methoxymethane | $3$. $142 \ pm$,$iii$. $108.9^o$ |
Difficult
View SolutionChoose the correct option by putting $T$ for the true statement and $F$ for the false statement.
$(i)$ Phenol is more acidic as compared to ethanol.
$(ii)$ Melting point of $o$-nitrophenol is lesser than $p$-nitrophenol.
$(iii)$ Neutralization of phenol is done with $NaHCO_3$.
$(iv)$ Nucleophilic substitution reaction is taking place in the aromatic ring of phenol.
$(i)$ Phenol is more acidic as compared to ethanol.
$(ii)$ Melting point of $o$-nitrophenol is lesser than $p$-nitrophenol.
$(iii)$ Neutralization of phenol is done with $NaHCO_3$.
$(iv)$ Nucleophilic substitution reaction is taking place in the aromatic ring of phenol.
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